Question

A population of values has a normal distribution with μ = 53.9 and σ = 17.9 . You intend to draw a random sample of size n = 28 . Find the probability that a single randomly selected value is between 57.6 and 58.3. P(57.6 < X < 58.3) = Find the probability that a sample of size n = 28 is randomly selected with a mean between 57.6 and 58.3. P(57.6 < M < 58.3) = Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

a)

Given,

= 53.9 , = 17.9

We convert this to standard normal as

P( X < x ) = P (Z < x - / )

So,

P(57.6 < X < 58.3) = P( X < 58.3) - P (X < 57.6)

= P( Z < 58.3 - 53.9 / 17.9) - P( Z < 57.6 - 53.9 / 17.9)

= P( Z< 0.246) - P (Z < 0.207)

= 0.5972 - 0.5820

= 0.0152

b)

Using central limit theorem,

P( M < x) = P( Z < x - / / sqrt(n) )

So,

P(57.6 < M < 58.3) = P( M < 58.3) - P (M < 57.6)

= P( Z < 58.3 - 53.9 / 17.9 / sqrt(28) ) - P( Z < 57.6 - 53.9 / 17.9 / sqrt(28) )

= P( Z < 1.301) - P( Z < 1.094)

= 0.9034 - 0.8630

= 0.0404